Let $Y_{i}$ be IID with $P(Y_{i}=1)$= $\frac{1}{2}=P(Y_{i}=-1)$. Define $K_{n}=\sum_{i=1}^{n}{Y_{i}}$.
Find the constant $r$ so that so that $$\Lambda_{n}=exp(K_{n} + rn)$$ is a Martingale.
I'm not sure that I have the right answer on this one. Any help will be appreciated.
The definition of a Martingale $$ \mathbb{E}\left[\Lambda_{n+1}|\mathcal{F}_{n}\right] = \Lambda_{n} $$ we have $$ \mathbb{E}\left[\mathrm{e}^{K_{n+1} + r(n+1)}|\mathcal{F}_{n}\right] = \mathbb{E}\left[\mathrm{e}^{Y_{n+1} + K_{n} + r(n+1)}|\mathcal{F}_{n}\right] $$ we can re-write the argument as $$ K_{n} + rn + Y_{n+1} + r $$ or $$ \mathbb{E}\left[\mathrm{e}^{K_{n+1} + r(n+1)}|\mathcal{F}_{n}\right] = \mathbb{E}\left[\Lambda_{n}\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right] = \Lambda_{n}\mathbb{E}\left[\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right] $$ we can extract the $Y_{n}$ since we know this up to the filtration time $n$. so we need to compute $$ \mathbb{E}\left[\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right] $$ For our condition to hold true we need $$ \mathbb{E}\left[\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right] = 1 $$ can you compute the expectation of the above? You have a simple calculation since we have discrete binary system!